Differentiability of multivariable functions represented through integral I have question about differentiability of functions from $\Bbb R^n$ to $\Bbb R$ represented through Lebesgue (Riemann) Integral. Which is somehow generalization of single variable case.
Question 1.
Let $\alpha : \Bbb R^n \to \Bbb (0, +\infty)$ be nice function (for example assume $C^1$), and Let $f : \Bbb R^n \to \Bbb R$ be continuous function   define $$ g(x) = \int_{B(x; \alpha(x))} f(y) ~ dy $$
is $g$ differentiable ? if so what is $\nabla g ?$
If not, what if I fix the center of the Ball  i.e.,
$$ g(x) = \int_{B(a; \alpha(x))} f(y) ~ dy  $$
Obviously  the answer is positive in dimension  one.
Question2. 
Let $f : \Bbb R^n \times \Bbb R^m \to \Bbb R $ be continuous and differentiable w.r.t to first argument,  (you even may assume $f$ is $C^1$ if I didn't make enough assumptions)
Let 
 $$ g(x) = \int_{ \Bbb R^m } f(x,y) ~ dy $$
Is $g$ differentiable ? if so what is $\nabla g ?$
 A: Question 2 Yes assume f and the partial derivatives D_i(f)  with respect to x are (Lebesgue) integrable in y and | D_i(f) (x,y) | <= G(y)  where G is integrable Then D_i(g) (x) = $\int (D_if(x,y) dy$  (integral is over R^n of course .  
A: The answer to Question 2 is no. On $\mathbb R^2$ define
$$f(x,y) = \frac{\sin^2(xy)}{1 +|y|^{3/2}}.$$
Then $f\in C^1(\mathbb R^2)$ and $f(x,y)$ is integrable in $y$ for each $x.$ Define $g$ as you did. Note $g(0)=0.$ Then for $x>0,$
$$\frac{g(x)-g(0)}{x-0} = \frac{1}{x}\int_{-\infty}^\infty\frac{\sin^2(xy)}{1 +|y|^{3/2}}\, dy.$$
Make the change of variables $y = t/x$ to get
$$\int_{-\infty}^\infty\frac{\sin^2(t)}{x^2 +|x|^{1/2}|t|^{3/2}}\, dy.$$
For $0<x<1,$ we have $x^2 < |x|^{1/2},$ so for such $x$ the last integral is greater than
$$\frac{1}{|x|^{1/2}}\int_{-\infty}^\infty\frac{\sin^2(t)}{1+|t|^{3/2}}\, dy.$$
As $x\to 0^+,$ the last expression $\to \infty.$ Thus $g'(0)$ does not exist.

Added later: For Question 1, we can do the following in the case where the center is fixed. Take $x=0$ for convenience. Then the "polar coordinates" formula for $\mathbb R^n$ shows that if $F(r) = \int_{B(0,r)} f(y)\,dy,$ then
$$F(r) = C_n\int_0^r \int_S f(s\zeta)\, d\sigma(\zeta)\, s^{n-1}\, ds.$$
Here $S$ is the unit sphere, $\sigma$ is normalized surface area measure on $S,$ and $C_n$ is the unnormalized surface area of $S.$ From the FTC, we get
$$F'(r) =C_n r^{n-1}\int_S f(r\zeta)\, d\sigma(\zeta).$$
Let $g(x)=F(\alpha (x)).$ From the above we can say
$$\frac{\partial g}{\partial x_k}(x) = F'(\alpha (x))\frac{\partial \alpha}{\partial x_k}(x) = C_n\alpha(x)^{n-1}\int_S f(\alpha(x)\zeta)\, d\sigma(\zeta)\cdot\frac{\partial \alpha}{\partial x_k}(x).$$
If we turn things around, letting the center $x$ move around while the radius remains constant, we get a problem I don't know to solve yet. There's an easy result if $f$ in $C^1.$ But I think you're right to focus on the case where $f$ is only continuous.
